Here you will learn angle of intersection of two curves formula with examples.
Let’s begin –
Angle of Intersection of Two Curves
The angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection.
Let C1 and C2 be two curves having equations y = f(x) and y = g(x) respectively.
and m1 = slope of tangent to y = f(x) at P = (dydx)C1
and m2 = slope of the tangent to y = g(x) at P = (dydx)C2
Angle between the curve is tanϕ = m_1 – m_2\over 1 + m_1 m_2
Orthogonal Curves
If the angle of two curves is at right angle, the two curves are equal to intersect orthogonally and the curves are called orthogonal curves.
If the curves are orthogonal then \phi = \pi\over 2
\therefore m_1 m_2 = -1
Note : Two curves ax^2 + by^2 = 1 and a’x^2 + b’y^2 = 1 will intersect orthogonally, if
1\over a – 1\over b = 1\over a’ – 1\over b’
Example : find the angle between the curves xy = 6 and x^2 y =12.
Solution : The equation of the two curves are
xy = 6 …….(i)
and, x^2 y = 12 …………(ii)
from (i) , we obtain y = 6\over x. Putting this value of y in (ii), we obtain
x^2 (6\over x) = 12 \implies 6x = 12
\implies x = 2
Putting x = 2 in (i) or (ii), we get y = 3.
Thus, the two curves intersect at P(2, 3).
Differentiating (i) with respect to x, we get
xdy\over dx + y = 0 \implies dy\over dx = -y\over x
\implies m_1 = ({dy\over dx})_{(2, 3)} = -3\over 2
Differentiating (ii) with respect to x, we get
x^2 dy\over dx + 2xy = 0 \implies dy\over dx = -2y\over x
\implies m_2 = ({dy\over dx})_{(2, 3)} = -3
Let \theta be the angle, then
tan \theta = m_1 – m_2\over 1 + m_1 m_2 = 3\over 11
\theta = tan^{-1} (3/11)
Related Questions
The angle of intersection between the curve x^2 = 32y and y^2 = 4x at point (16, 8) is